A spring of force constant k is cut into two parts such that length of one part is double to another part. What will be the force constant of larger part?

A spring of force constant k is cut into two parts such that length of

1.	A spring of force constant k is cut into two parts such that length of one part is double to another part. What will be the force constant of larger part?
Answer» `(2)/(3)k` `(3)/(2)k` `3k` `6k` Solution :FORCE CONSTANT of SPRING is inversely PROPORTIONAL to its length, `therefore k propto (1)/(l)` `therefore kl = 1"""…….."(1)` Given that `l_(1)= 2l_(2)` and `l= l_(1)+l_(2)` `=2l_(2)+l_(2)` `l= 3l_(2)` and `l= l_(1)+(l_1)/(2)= (3l_1)/(2)"""........."(2)` Now, `k_(1)l_(1)= k_(2)l_(2) = kl` `therefore k_(1)l_(1) = kl` `k_(1)l_(1)= k xx (3l_1)/(2)` [From equation (1)] `therefore k_(1)= (3)/(2)k`.